Let H and K be subgroup of an abelian group G (not neccessarily finite). Suppose order of H and K are a and b respectively. Prove that there exists a subgroup with order L, where L = lcm(a,b).

I tried to use the Lagrange theorem but only manage to prove it for the case when g is finite. I tried to use the product formula: |HK|/|H| = |K|/|H intersect K|. Sub a and b in, i obtain: |HK||H intersect K| = ab.

Hence, it suffices to show that |H intersect K| = gcd (a,b) since lcm(a,b)*gcd(a,b) = ab. But how to show that? or is there anothe method to solve this?

Thank You.

Nov 5th 2011, 06:20 AM

Tinyboss

Re: Abelian Group

Have you encountered the fundamental theorem of finitely-generated abelian groups yet?

Nov 5th 2011, 01:37 PM

Deveno

Re: Abelian Group

this looks like that problem from herstein...the one real early in the book, before the structure theorem for abelian groups, or even the sylow theorems. an "elementary proof" isn't easy...